IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

swan1nas

#1

if two distinct chords of a parabola y²= 4ax , passing thru (a, 2a) are bisected by the line x+y=1, then the length of latus rectumcan be???

A) 2

B) 1

C) 4

D) none

#2

if the normals drawn at the end points of a variable chord PQ of the parabola y²= 4ax intersect at parabola, then the locus of the point of intersection of the tangent drawn at the points P and Q is :

A) x+a=0

B) x-2a=0

C) y² - 4x +6 =0

D) none

#3

the equation of shortest distance between the parabolaa y² = 4x and the circle x² + y² - 4x - 2y + 4=0 is

A) x+y=3

B) x-y=3

C) 2x+y=5

D) none

#4

the length of shortest normal chord of the parabola y² = 4ax is of length

A) 2a

3

B) 4a

3

C) 6a

3

D) 8a

3

#5

the normals at the ends of the latus rectum of the parabola y²= 4ax meet the parabola again at P and P'. then PP' is equal to

A) a

B) 2a
C) 12a

D) 3a

#6

If the line y-

3x + 3 = 0 cuts the parabola y² = x + 2 at A and B, and if P = (

3, 0), then PA . PB is equal to

A) 2(

3+2)/3

B) 4

3 /2

C) 4 ( 2-

3) / 3

D) 4(

3 +2)/3

#7

If the normal to the parabola y²= 4ax at the point (at², 2at) cuts the parabola again at (aT², 2aT) then

A) T²

8

B) T

( -

, -8) U ( 8 ,

)

C) -2

T

2

D) T² < 8

#8

The normals at three points P, Q, R of the parabola y²= 4ax meet in (h,k). The centroid of the triangle POR lies on

A) x=0

B) y=0

C) x= (-a)

D) y=a

#9

The normal at the point P( ap², 2ap) meets the parabola y²= 4ax again at Q(aq², 2aq) such that the lines joining the origin to P and Q are at right angle. Then

A) p² = 2

B) q² = 2

C) p= 2q

D) q= 2p

#10

If the chord of contact of tangents froma point P(h,k) to the cirlce X²+ Y²= A² touches the circle X² + (Y-A)² = A², then what is the locus of P?

#11

The length of the chord of the parabola X²= 4Y passing through the vertex and having slpoe cotA is

A) 4cosAcosec²A

B) 4tanAseca

c) 4sinAsec²A

D) none

#12

If the normals at the end points of a variable chord PQ of the parabola y²- 4y - 2x = 0 are perpendicular, then the tangents at P and Q will intersect at

A) x+y=3

B) 3x-7=0

C) y+3=0

D) 2x + 5=0

vpunithreddy

u can post only one q at a time

i'm solvin 1st q

letz solve dis by elimination of options

substitute x=1-y in da parabola

solve da quadratic in y u have 2 get a positive soln and a negative soln if u substitute the value of a frm options.a&b cant be possible.

so 4 c u get 2 values one positive and negative

and by goemetry u can get da the 2 chords r latus rectum and a perpendicular 2 da given line

so da length of latus rectum must be 4

hope i'm right

akku

#3

A)x+y=3

the perpendicular frm the parabola 2 the circle will b the line of shortest distance .As its pptothe circle it will pass thr its centre .The coordinates of centre(2,1) satisfy a)HENCE the answer

akku

#12

D)

pt of intersectn of pp tangents lie on the directrix

rewriting the eqn of the parabola

y^2-4y+4=2x+4

=> (y-2)^2=2(x+2)

Y^2=4aX

DIRECTRIX x=-a

x+2= -1/2 (eqn of directx comp wid standard form)

=> 2x+5=0 ans

#11

A)

Eqn of chord y=xcotA (c=0 since it passes thr vertex(0,0)

the other end of chord(4cotA,4cot^2A)

length of chord:-

(16cot^2A+16cot^4A=4cotA

1+cot^A=4cotAcosecA=4cosAcosec^2A

catch_arnnie

#10

If the chord of contact of tangents froma point P(h,k) to the cirlce X²+ Y²= A² touches the circle X² + (Y-A)² = A², then what is the locus of P?

ans::

equation of chord of contact --> hx + ky - A² = 0

since, the chord of contact touches the circle X² + (Y-A)² = A², (center=(0,A), radius = A)

therefore, its distance from the center = to the radius of circle

=> | (Ak - A² )/

(h² + k²) | = A

squaring both sides, & solving

=> (k-A)²/(h² + k² ) = 1

=> A² - 2Ak = h²

=> x² + 2Ay - A² = 0 -----> ans.

i guess this should be correct..

catch_arnnie

#9

The normal at the point P( ap², 2ap) meets the parabola y²= 4ax again at Q(aq², 2aq) such that the lines joining the origin to P and Q are at right angle. Then

A) p² = 2

B) q² = 2

C) p= 2q

D) q= 2p

ans:: A)

solution::

since, the lines joining the origin to P and Q are at right angle,

therefore, (ap²/ 2ap)*(aq²/2aq) = -1 [product of slopes = -1]

=>4/pq = -1 => q = -4/p

now, here i'm using a result whose proof i'm not giving rite now, but if you want it then i'll surely give it to you.

whenever a normal from one point on the parabola having coordinates like (at^2 , 2at) , intersects the parabola at some point (at'^2,2at'), then t' = -t-2/t

using this, q = -p - 2/p

putting the value of q from above

-4/p = - p - 2/p

=> p² = 2

hence, option A) is correct

..

sankydreams

# 5

The ends of latus rectum => (a,2a),(a,-2a)

For (a,2a), t=1

Normal at this pt wud meet parabola again at point where

t' =-t-2/t

Putting t=1

t' = -3

Similarly t'' for (a,-2a) = 3

Coordinates of P=>(at'^2,2at') = (9a,-6a)

Coordinates of P'=>(at''^2,2at'') =(9a,6a)

Clearly PP' = 12a

Hence (C)

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sankydreams

#7

Since normal at pt t meets the curve again at pt T

T= - t - 2/t

This equation becomes a quadratic in t => t^2 + tT + 2 =0

Roots of this quadratic eqn must be real

=> 0

T^2 - 8

=> T^2

8

HenceA

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